Every matrix $A\in SL_2(\mathbb{Z})$ induces a self homeomorphism of $S^1\times S^1=\mathbb{R}^2/\mathbb{Z}^2$. For different matrices these homeomorphisms are not homotopic, as the induced map on $\pi_1(S_1\times S_1)$ is given by $A$ (w.r.t the induced basis).

So I am wondering, whether a similar construction also works for other spheres than $S^1$. So is there any non-obvious self-homeomorphism of $S^2\times S^2$ (I know i can use degree $-1$ maps in any choice of coordinates / flip the coordinates /compose such maps).

There are some more coming from the fact that the Grassmannian of oriented 2-planes in 4-dimensions is diffeomorphic to $S^2$x$S^2$. Thus $S^2$x$S^2$ is also $SO(4)/(SO(2)\times SO(2))$.
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j.c.Oct 11 '10 at 15:38

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The good question might be: "what can be said about the homotopy type of Diff($S^2\times S^2$)?"
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André HenriquesOct 11 '10 at 15:56

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@Andre. That's a fine question, but sadly we don't have a single computation of $\pi_i (Diff(X))$ for a closed 4-manifold $X$ and an integer $i \geq 0$.
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Tim PerutzOct 11 '10 at 17:31

Which finite groups can act freely on a product of two 2-spheres? Are there any besides the obvious Klein 4-group?
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Robert BellOct 11 '10 at 18:47

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rbell - No. Any such action would map to $\text{Aut}(H^*(S^2 \times S^2))$, which is the 4-group. The kernel would act trivially on homology and would therefore have fixed points by the Lefschetz Fixed-Point Theorem.
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Greg KuperbergOct 11 '10 at 18:58

2 Answers
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The matrix $\text{SL}(2,\mathbb{Z})$ acts on $H^n(S^n \times S^n)$; one interpretation of your question is whether this action lifts to $\text{Diff}(S^n \times S^n)$. There is a simple reason that it doesn't when $n$ is even: The intersection form on $H^n(S^n \times S^n)$ is symmetric rather than anti-symmetric, and any diffeomorphism has to preserve this form. This more or less nails down everything, in the weaker category of homotopy self-equivalences; in this setting you can't do anything other than exchange the spheres or apply degree $-1$ maps. (But the full homotopy structure of the $\text{Diff}(S^n \times S^n)$ could be much more complicated.)

When $n$ is odd things are much more complicated. If $n=3$ or $n=7$, you can use multiplication in the unit quaternions or the unit octonions to lift the matrix
$$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$
and its transpose. These matrices generate $\text{SL}(2,\mathbb{Z})$. On the other hand, for any other value of $n$, there is no diffeomorphism, nor even any homotopy equivalence, that realizes this matrix. Because, if you composed such a map with projection onto one of the factors, it would turn $S^n$ into an H-space.

I don't know of a way to prove more than that just by citation for some other large, odd value of $n$. In other words, I know that you can't get all of $\text{SL}(2,\mathbb{Z})$, but I don't know how big of a group you can get.

A small comment extending Greg's comments, given an element of $[g] \in \pi_n Diff(S^n)$ you can construct a diffeomorphism

$f : S^n \times S^n \to S^n \times S^n$ by sending the pair $(x,y)$ to $(x,g_x(y))$, so the corresponding matrices would be upper-triangular as in Greg's examples.

$Diff(S^n)$ has the homotopy-type of $O_{n+1} \times Diff(D^n)$.

Greg's examples come from situations where there are non-trivial elements of $\pi_n O_{n+1}$ that you "know" because they're coming from splittings of the fibrations $SO_n \to SO_{n+1} \to S^n$, but you could also get non-trivial elements from finding elements in $\pi_n Diff(D^n)$. My understanding is there's little to nothing known about these homotopy groups. The rational homotopy $\pi_j Diff(D^n)\otimes \mathbb Q$ of $Diff(D^n)$ is only understood in the "Igusa Stable Range" of $j \leq \min\{ \frac{n-4}{3}, \frac{n-7}{2} \}$ so it's not useful for the kind of constructions you'd like.